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Schach [20]
3 years ago
9

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

end of the spring, which stretches to a new length of 37 cm. What is the spring constant, k, in N/m?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

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Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

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7 0
3 years ago
The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th
ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, B=2\ mT=2.2\times 10^{-3}\ T

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Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :                  

B=\dfrac{\mu_oNI}{2\pi r}

r=\dfrac{\mu_oNI}{2\pi B}  

r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}  

r = 0.00523 m

or

r=5.23\times 10^{-3}\ m

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0
3 years ago
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Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
kkurt [141]

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

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Please help me with this
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Answer:

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